Re: Everyone on this forum besides Keith has been a damned liar about this point

On 6/10/2025 6:59 AM, Mikko wrote:

On 2025-06-09 14:38:02 +0000, olcott said:
 

On 6/9/2025 2:56 AM, Mikko wrote:

On 2025-06-09 02:50:59 +0000, olcott said:
>

void DDD()
{
   HHH(DDD);
   return;
}
>
The *input* to simulating termination analyzer HHH(DDD)
specifies recursive simulation that can never reach its
*simulated "return" instruction final halt state*
>
*Every rebuttal to this changes the words*

>
Being called a "liar" by a liar does not damn.
>
As is clear from the above C code, DDD() specifies what HHH specifies
for the case it is called with DDD as the only argument. In particular,
if HHH specifies a recursive for that case then so does DDD. And if
HHH specifies a recursive simulation that can never reach its final
halt state then so does DDD. And if HHH specifies a non-halting
behaviour so does DDD. Etc.

>
That is not quite the way that it actually works.

 Yes it is. If it were not you would have pointed where there is an
error.
 

I only point out the first error and the skip the rest of the post.
I usually have to point out the same error dozens of times before
anyone notices that I said anything at all. That is why I skip the
rest of the post after the first error.
*You can find your pwn error using this code*
https://github.com/plolcott/x86utm/blob/master/Halt7.c

<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its
     input D until H correctly determines that its simulated D
     would never stop running unless aborted then
>
     H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>
DDD emulated by HHH, the directly executed DDD()
and the directly executed HHH() would never stop
running unless HHH aborts its simulation of DDD.

 A counterfactual hypothesis is not a valid substitute of the actual
input as the basis of determination.
 

None of that is counter-factual. That you say it is
proves your own ignorance.

*According to the above criteria that means*
>
When DDD does abort its simulation of DDD then
HHH correctly reports that its input specifies
a non-halting sequence of configurations.

 No, it does not, as HHH did not correctly determine that the rest
of the behaviour is non-terminating. If the simulated HHH were
similated further it would abort its simulation and return.

This is woeful ignorance.
I challenge you to provide all of the details
of this so that your error becomes more clear to you.
void DDD()
{
   HHH(DDD);
   return;
}
HHH is invoked and simulates DDD that calls HHH(DDD)
simulated HHH simulates DDD that calls HHH(DDD)
simulated HHH simulates DDD that calls HHH(DDD)
simulated HHH simulates DDD that calls HHH(DDD)
simulated HHH simulates DDD that calls HHH(DDD)
simulated HHH simulates DDD that calls HHH(DDD)...
In a TM this would repeat forever.

The
actual input specifies a partial simulation and later halting.
 

Unless the simulated DDD reaches its own simulated "return"
statement final halt state this simulated DDD NEVER HALTS.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer